# What is the perimeter of a triangle with corners at (3 ,6 ), (1 ,5 ), and (8 ,1 )?

Mar 16, 2016

Perimeter, $P = \overline{A B} + \overline{B C} + \overline{C A} = \sqrt{5} + \sqrt{65} + 5 \sqrt{2}$

#### Explanation:

Use the distance formula to determine: $\overline{A B} , \overline{B C} , \overline{C A}$
The Distance Formula between 2 points:
${P}_{1} \left({x}_{1} , {y}_{1}\right)$, ${P}_{2} \left({x}_{2} , {y}_{2}\right)$
$\overline{{P}_{1} {P}_{2}} = s q t \left({\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}\right)$ so then
$\overline{A B} = \sqrt{{\left(3 - 1\right)}^{2} + {\left(6 - 5\right)}^{2}} = \sqrt{5}$
$\overline{B C} = \sqrt{{\left(1 - 8\right)}^{2} + {\left(5 - 1\right)}^{2}} = \sqrt{65}$
$\overline{C A} = \sqrt{{\left(8 - 3\right)}^{2} + {\left(1 - 6\right)}^{2}} = 5 \sqrt{2}$
Perimeter, $P = \overline{A B} + \overline{B C} + \overline{C A} = \sqrt{5} + \sqrt{65} + 5 \sqrt{2}$